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16 March, 00:14

A mixture of 0.438 M H2, 0.444 M I2, and 0.895 M HI is enclosed in a vessel and heated to 430 °C. H2 (g) + I2 (g) 2 HI (g) Kc = 54.3 at 430∘C Calculate the equilibrium concentrations of each gas at 430∘C.

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  1. 16 March, 00:34
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    [H₂] = 0.178 M

    [I₂] = 0.184 M

    [HI] = 1.415 M

    Explanation:

    For the equilibrium:

    H₂ (g) + I₂ (g) ⇄ 2 HI (g)

    the equilibrium constant is given by the equation:

    Kc = [ HI]² / [H₂][I₂]

    Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i. e towards reactants or product side.

    Q = (0.895) ² / (0.438) (0.444) = 4.12

    Q is less than Kc so the reaction will favor the product side.

    We can set up the following table to account for all the species at equilibrium:

    H₂ I₂ HI

    initial 0.438 0.444 0.895

    change - x - x + 2x

    equilibrium 0.438 - x 0.444 - x 0.895 + 2x

    Now we are in position to express these concentrations in terms of the equilibrium conctant, Kc

    54.3 = (0.895 + 2x) ² / (0.438 - x) (0.444 - x)

    performing the calculatiopns will result in a quadratic equation:

    0.801 + 3.580x + 4x² = (0.194 - 0.882x + x²) x 54.3

    Upon rearrangement and some algebra, we have

    0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²

    0 = 9.733 - 51.473 x + 54.3 x²

    This equation has two roots X₁ = 0.687 and X₂ = 0.26

    The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.

    Therefore the concentrations at equilibrium of each gas are:

    [H₂] = (0.438 - 0.260) = 0.178 M

    [I₂] = (0.444 - 0.260) M = 0.184 M

    [HI] = [0.895 + 2x (0.260) ] M = 1.415 M

    Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.
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